Only one because there is one variable and the combined coefficients of x are non-zero.
(0.75(x+40)=0.70(x+20), 0.05x=14-30=-16, x=-320)
Answer:
A
Step-by-step explanation:
beacause if you try to multiple 4x7.5 divided by 2 try it
Answer:
0.000036
Explanation:
When multiplying a number by 10 to the power of a positive exponent (e.g.
10
3
), move the decimal in the number that many spaces to the right (e.g.
4.5
⋅
10
3
=
4500
).
When multiplying a number by 10 to the power of a negative exponent (e.g.
10
−
4
), move the decimal in the number (the absolute value of) that many spaces to the left (e.g.
4.5
⋅
10
−
4
=
0.00045
).
For
3.6
⋅
10
−
5
, move the decimal five spaces to the left:
0.000036
Step-by-step explanation:
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.
So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))
now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16
Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84
So we have our linear approximation for the two.
If you wanted to, you could check your answer by finding g(x). Since you know g'(x), take the antiderivative and we will get
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35
So just to check our linear approximations use that to find g(2.99) and g(3.01)
g(2.99) = -5.1597
g(3.01) = -4.8397
So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer. Not a bad method if you ever need to use it.
Answer:
5/4k^2
Step-by-step explanation:
P=5\dfrac{k}{6}\times \dfrac{3}{2k^3}.
We will be using the following property of exponents:
\dfrac{a^x}{a^y}=a^{x-y}.
We have
P\\\\\\=5\dfrac{k}{6}\times\dfrac{3}{2k^3}\\\\\\=\dfrac{5}{6}\times\dfrac{3}{2}k^{1-3}\\\\\\=\dfrac{5}{4}k^{-2}=\dfrac{5}{4k^2}.
Thus, the required product is \dfrac{5}{4k^2}.
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